Problem: Factor the quadratic expression completely. $2x^2-13x+20=$
Solution: Since the terms in the expression do not share a common monomial factor and the coefficient on the leading $x^2$ term is not $1$, let's factor by grouping. The expression ${2}x^2{-13}x{+20}$ is in the form ${A}x^2+{B}x+{C}$. First, we need to find two integers ${a}$ and ${b}$ such that: $\begin{cases} &{a}+{b}={B}={-13} \\\\ &{ab}={A}{C}= ({2})({20})=40 \end{cases}$ We find that ${a}={-5}$ and ${b}={-8}$ satisfy these conditions, since ${-5}+({-8})={-13}$ and $({-5})({-8})=40$. Next, we can use these values to rewrite the $x$ -term and factor by grouping. $\begin{aligned} 2x^2-13x+20&=2x^2{-8}x{-5}x+20 \\\\ &=2x(x-4)-5(x-4) \\\\ &=(x-4)(2x-5) \end{aligned}$ In conclusion, $2x^2-13x+20=(x-4)(2x-5)$